Lorentz Force Law

Equivalent forms

F = qvB\sin\theta \quad \text{(magnetic only)}

\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}

r = \frac{mv}{qB} \quad \text{(cyclotron radius)}

One equation governs every charged particle in the universe — from electrons in wires to cosmic rays spiraling through galaxies.

Symbol	Name	SI unit	Dimension	Range
$F$	Lorentz forceoutput Total electromagnetic force on the charged particle	N	M·L·T⁻²	0 – 1e-10
$q$	Charge Electric charge of the particle	C	I·T	-1e-17 – 1e-17
$v$	Velocity Speed of the charged particle	m/s	L·T⁻¹	0 – 299792458
$B$	Magnetic field External magnetic field strength	T	M·T⁻²·I⁻¹	0 – 10
$E$	Electric field External electric field strength	V/m	M·L·T⁻³·I⁻¹	0 – 1000000

Unit systems

Symbol	SI	CGS	Imperial
$F$	N	dyn	lbf
$q$	C	statC	C
$v$	m/s	cm/s	ft/s
$B$	T	G	T
$E$	V/m	statV/cm	V/m

Where it holds

Valid for point charges at all speeds when combined with relativistic momentum. The non-relativistic form

(F = ma)

applies for v << c. Does not account for radiation reaction — for that, use the Abraham–Lorentz force.

Dimensional analysis

$[F] = [q]([E] + [v]\cdot [B]) \to C(V/m + m/s\cdot T) = C(N/C + N/C) = N \checkmark$

Discovery

Hendrik Lorentz · 1895

Lorentz unified the electric and magnetic forces on a moving charge into a single expression, providing the bridge between Maxwell's field equations and Newton's mechanics of particles.

Try this

Why do charged particles spiral in the Northern Lights instead of flying straight?

An electron moving at 2 × 10⁶ m/s enters a 0.01 T magnetic field perpendicular to its velocity. Find the radius of its circular orbit.

Research status: stable

Real-world applications

Cyclotrons and synchrotrons in particle physics research
Mass spectrometers for chemical and isotope analysis
Hall effect sensors in automotive and industrial applications
Aurora borealis — charged solar wind particles spiral along Earth's magnetic field

Common misconceptions

The magnetic force does NO work — it changes direction but not speed
The force is perpendicular to v, not to B (it's perpendicular to both via the cross product)
The 'magnetic force' between two wires is really a relativistic electric effect in the electrons' rest frame

Experimental verification

Thomson's cathode ray experiments (1897) measured e/m using crossed E and B fields. Cyclotron resonance, mass spectrometry, and Hall effect measurements all confirm the Lorentz force to high precision.

Derivation

Postulated by Lorentz as the force on a charge in given E and B fields.

Consistent with special relativity: the electric and magnetic forces transform into each other under Lorentz boosts.

The magnetic force is a relativistic correction to the electric force between moving charges.

Limiting cases

v = 0

⟶

F = qE

(electric force only)A stationary charge feels no magnetic force — only electric fields act on it.

E = 0

⟶

F = qvB \sin \theta

Pure magnetic force — always perpendicular to velocity, so it does no work.

v ∥ B⟶ F_magnetic

= 0

Cross product vanishes for parallel vectors — the particle moves in a straight line along B.

What if…

What if the particle is a proton instead of an electron?

$r = m_p \times v/(qB) = 1.673\times 10^{-27} \times 2\times 10^{6} / (1.602\times 10^{-19} \times 0.01) = 2.09\,\mathrm{m}$ — about $1836\times$ larger than the electron orbit.

What if the velocity is parallel to B?

No magnetic force (cross product $= 0)$ . The particle travels in a straight line along B — no deflection at all.

What if both E and B are present and perpendicular?

Particle drifts perpendicular to both E and B with velocity v_drift $= E\times B/B^{2}$ . This is the $E\times B$ drift, fundamental in plasma physics.

1

Cyclotron radius of an electron

Given ·

v:: 2000000
B:: 0.01
m:: 9.1093837015e-31
q:: 1.602176634e-19

Find · Radius of circular orbit

Steps

Magnetic force provides centripetal acceleration: $qvB = mv^{2}/r$
Solve for r: $r = mv/(qB)$
Substitute: $r = (9.1093837015\times 10^{-31} \times 2\times 10^{6}) / (1.602176634\times 10^{-19} \times 0.01)$
Numerator: $1.8219\times 10^{-24} kg\cdot m/s$
Denominator: $1.602\times 10^{-21} N\cdot s/m \to r = 1.138\times 10^{-3} m \approx 1.14\,\mathrm{mm}$

Result ·

r = mv/(qB) = 9.109\times 10^{-31} \times 2\times 10^{6} / (1.602\times 10^{-19} \times 0.01) = 1.138\times 10^{-3} m \approx 1.14\,\mathrm{mm}

2

Velocity selector (crossed E and B fields)

Given ·

E field:: 100000
B:: 0.05

Find · Selected velocity

Steps

Electric force: $F_E = qE (say$ , upward)
Magnetic force: $F_B = qvB (say$ , downward for the right velocity)
For straight-through passage: $qE = qvB$
Cancel q: $v = E/B = 10^{5} / 0.05 = 2\times 10^{6} m/s$
Only particles with this exact speed pass through undeflected — independent of charge and mass

Result ·

v = E/B = 10^{5} / 0.05 = 2\times 10^{6} m/s

Coulomb's Law→Biot–Savart Law→Cyclotron Frequency→